Cho các dãy số (un), (vn) với u n = √ 4 n 2 + 5 n + 1 và vn = 2n + 1

a) Ta có v_n = 2n + 1 \( = n\left( {2 + \frac{1}{n}} \right)\).

Vì \(\mathop {\lim }\limits_{n \to + \infty } n = + \infty ;\mathop {\lim }\limits_{n \to + \infty } \left( {2 + \frac{1}{n}} \right) = 2\). Do đó \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = \mathop {\lim }\limits_{n \to + \infty } \left( {2n + 1} \right) = + \infty \).

b) Ta có \({u_n} = \sqrt {4{n^2} + 5n + 1} = n\sqrt {4 + \frac{5}{n} + \frac{1}{{{n^2}}}} \).

Vì \(\mathop {\lim }\limits_{n \to + \infty } n = + \infty ;\mathop {\lim }\limits_{n \to + \infty } \sqrt {4 + \frac{5}{n} + \frac{1}{{{n^2}}}} = 2\) nên \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = \mathop {\lim }\limits_{n \to + \infty } \sqrt {4{n^2} + 5n + 1} = + \infty \).

c) Có \({u_n} + {v_n} = \sqrt {4{n^2} + 5n + 1} + 2n + 1 = n\left( {\sqrt {4 + \frac{5}{n} + \frac{1}{{{n^2}}}} + 2 + \frac{1}{n}} \right)\).

Vì \(\mathop {\lim }\limits_{n \to + \infty } n = + \infty ;\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {4 + \frac{5}{n} + \frac{1}{{{n^2}}}} + 2 + \frac{1}{n}} \right) = 4\).

Do đó \(\mathop {\lim }\limits_{n \to + \infty } \left( {{u_n} + {v_n}} \right) = + \infty \).

d) Có \({u_n} - {v_n} = \sqrt {4{n^2} + 5n + 1} - \left( {2n + 1} \right) = \frac{{{{\left( {\sqrt {4{n^2} + 5n + 1} } \right)}^2} - {{\left( {2n + 1} \right)}^2}}}{{\sqrt {4{n^2} + 5n + 1} + \left( {2n + 1} \right)}} = \frac{n}{{\sqrt {4{n^2} + 5n + 1} + \left( {2n + 1} \right)}}\).

Do đó \(\mathop {\lim }\limits_{n \to + \infty } \left( {{u_n} - {v_n}} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{n}{{\sqrt {4{n^2} + 5n + 1} + \left( {2n + 1} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{\sqrt {4 + \frac{5}{n} + \frac{1}{{{n^2}}}} + \left( {2 + \frac{1}{n}} \right)}} = \frac{1}{4}\).

Đáp án: a) Sai;    b) Đúng;    c) Sai;    d) Đúng.